Count Alternating Subarrays

Count Alternating Subarrays - Problem

Array Math Medium

You are given a binary array nums. We call a subarray alternating if no two adjacent elements in the subarray have the same value.

Return the number of alternating subarrays in nums.

A subarray is a contiguous non-empty sequence of elements within an array.

Input & Output

Example 1 — Fully Alternating

$ Input: nums = [0,1,0,1]

› Output: 10

💡 Note: All subarrays are alternating: [0], [1], [0], [1], [0,1], [1,0], [0,1], [0,1,0], [1,0,1], [0,1,0,1]. Total = 10.

Example 2 — Mixed Pattern

$ Input: nums = [0,1,1,0]

› Output: 5

💡 Note: Alternating subarrays: [0], [1], [1], [0], [0,1]. The segment [1,1] breaks alternation, so [0,1,1] and longer are not valid.

Example 3 — All Same

$ Input: nums = [1,1,1]

› Output: 3

💡 Note: Only single-element subarrays are alternating: [1], [1], [1]. No multi-element alternating subarrays exist.

Constraints

1 ≤ nums.length ≤ 10⁵
nums[i] is either 0 or 1

Visualization

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Asked in

G Google 25 a Amazon 20 M Microsoft 15

The key insight is that each alternating segment of length n contributes n*(n+1)/2 subarrays. Best approach uses sliding window to identify segments in O(n) time. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(n³) Space: O(1)

Generate all possible subarrays and check each one to see if adjacent elements are different. Count how many satisfy the alternating property.

Sliding Window Optimization

⏱️ Time: O(n) Space: O(1)

Instead of checking every subarray individually, find continuous alternating segments and calculate how many subarrays each segment contributes using the formula n*(n+1)/2.

Brute Force — Algorithm Steps

Generate all possible subarrays
Check each subarray for alternating pattern
Count valid alternating subarrays

Visualization

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Step-by-Step Walkthrough

Generate Subarrays

Create all possible contiguous subsequences

Check Pattern

Verify each subarray has alternating elements

Count Valid

Sum up all alternating subarrays

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize) {
    int count = 0;
    
    // Check all possible subarrays
    for (int i = 0; i < numsSize; i++) {
        for (int j = i; j < numsSize; j++) {
            // Check if subarray [i:j+1] is alternating
            int isAlternating = 1;
            for (int k = i; k < j; k++) {
                if (nums[k] == nums[k + 1]) {
                    isAlternating = 0;
                    break;
                }
            }
            if (isAlternating) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int size = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, size);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

O(n²) subarrays times O(n) to check each one

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

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Code -

Time & Space Complexity

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